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Arctan problem help

  1. Apr 30, 2005 #1
    [tex]y=tan^{-1}\left(\frac{1}{ln(x)}\right)[/tex]
    [tex]y'=\frac{1}{1-\left(\frac{1}{ln(x)}\right)^2}[/tex]
    [tex]y'=\frac{1}{1-(lnx)^{-2}}[/tex]

    is this correct?

    do I have to take into account the derivative of [itex]\frac{1}{ln(x)}[/itex]?

    if I do, what is the derivative of [itex]\frac{1}{ln(x)}[/itex]?
     
    Last edited: Apr 30, 2005
  2. jcsd
  3. Apr 30, 2005 #2

    Pyrrhus

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    Yes, you need to take into account the derivative of [itex] \frac{1}{\ln(x)} [/itex].

    You could either do the derivative of a division or do the derivative of [itex] (\ln(x))^{-1} [/itex].
     
  4. Apr 30, 2005 #3
    [tex]y'=\frac{1}{1-(lnx)^{-2}}*\frac{1}{x(lnx^2)}[/tex]
    [tex]y'=\frac{1}{x(lnx^2)-x}}[/tex]

    is that it?
     
  5. Apr 30, 2005 #4

    Pyrrhus

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    Little changes...

    [tex]y'=\frac{1}{1+(lnx)^{-2}}*\frac{-1}{x(\ln(x))^{2}}[/tex]

    Something else i noticed

    [tex] (tan^{-1} (f(x)))' = \frac{f'(x)}{1+(f(x))^2} [/tex]
     
    Last edited: Apr 30, 2005
  6. Apr 30, 2005 #5
    can that become [tex]y'=\frac{1}{-x(lnx^2)+x}}[/tex]?
     
  7. Apr 30, 2005 #6

    Pyrrhus

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    Stop putting the square next to the x like that, use parentheses.
     
  8. Apr 30, 2005 #7
    sorry, didn't notice that

    [tex]y'=\frac{1}{-x(lnx)^2+x}}[/tex]

    that's what I meant
     
  9. Apr 30, 2005 #8

    Pyrrhus

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    Sure, except for the sign problem.
     
  10. Apr 30, 2005 #9
    [tex]y'=\frac{1}{-x\ln(x)^{2}-x}[/tex]
     
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